Question

An approximation for \(n !,\) when \(n\) is large, is given by $$ n ! \approx \sqrt{2 n \pi}\left(\frac{n}{e}\right)^{n}\left(1+\frac{1}{12 n-1}\right) $$ Calculate \(12 !, 20 !,\) and \(25 !\) on your calculator. Then use Stirling's formula to approximate \(12 !, 20 !,\) and \(25 !\)

Short answer

12! ≈ 4.79 x 10^8, 20! ≈ 2.43 x 10^18, 25! ≈ 1.54 x 10^25 using Stirling's formula.

Step-by-step solution

- Calculate 12!, 20!, and 25! using a calculator

Use a calculator to find the values of 12!, 20!, and 25! The factorial function is typically available on scientific calculators or can be computed using online tools.

Values obtained from Calculator

12! = 479001600, 20! = 2.432902e18 (2.43290200817664 × 10^{18}), 25! = 1.551121e25 (1.5511210043331 × 10^{25})

- Stirling's Approximation Formula

Stirling's approximation is given by: \[ n! \approx \sqrt{2 n \pi}\left(\frac{n}{e}\right)^{n}\left(1+\frac{1}{12n-1}\right) \]

- Calculate 12! using Stirling's Formula

Substitute \(n = 12\) into the formula: \[ 12! \approx \sqrt{2 \cdot 12 \cdot \pi} \left( \frac{12}{e} \right)^{12} \left( 1 + \frac{1}{12 \cdot 12 - 1} \right) \]. Simplify inside the parentheses first: \[ 12! \approx \sqrt{24 \pi} \left( \frac{12}{e} \right)^{12} \left(1 + \frac{1}{143} \right) \] Calculate the numerical values to find the approximate answer.

- Calculate 20! using Stirling's Formula

Substitute \(n = 20\) into the formula: \[ 20! \approx \sqrt{2 \cdot 20 \cdot \pi} \left( \frac{20}{e} \right)^{20} \left( 1 + \frac{1}{12 \cdot 20 - 1} \right) \]. Simplify and calculate the numerical values to find the approximate answer.

- Calculate 25! using Stirling's Formula

Substitute \(n = 25\) into the formula: \[ 25! \approx \sqrt{2 \cdot 25 \cdot \pi} \left( \frac{25}{e} \right)^{25} \left(1 + \frac{1}{12 \cdot 25 - 1} \right) \]. Simplify and calculate the numerical values to find the approximate answer.

- Compare Results

Compare the results obtained using Stirling's approximation with the calculated values from the calculator in Step 1.

Key concepts

Factorial

A factorial, denoted as \(n!\), is the product of all positive integers from 1 to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow very fast as \(n\) increases. Even for moderate \(n\), the resultant product becomes extremely large.
Factorials are crucial in various mathematical fields such as combinatorics, number theory, and calculus.

• **Permutations:** Calculating different ways to arrange items
• **Combinations:** Computing ways to select items from a set Without replacements

Because they grow so rapidly, calculating large factorials directly is often impractical without the assistance of special techniques or tools like Stirling's approximation.

Approximations in Mathematics

Approximations in mathematics are used to find simpler expressions or values that are close to the exact result. Stirling's approximation, for example, approximates factorials: \[ n! \approx \sqrt{2n\pi} \left( \frac{n}{e} \right)^n \left(1 + \frac{1}{12n - 1} \right). \]It simplifies the computation of large factorials.
Here's what each part of the formula means:

• \(\sqrt{2n\pi}\): A scaling factor that adjusts the magnitude
• \(\left( \frac{n}{e} \right)^n\): A term that captures the rapid growth
• \(1 + \frac{1}{12n - 1}\): A small correction factor improving accuracy for finite \(n\)

Approximations are particularly useful when exact solutions are difficult or impossible to find. They balance simplicity and accuracy, providing insights into the behavior of numbers and functions.

Large Numbers in Computations

Handling large numbers in computations can be challenging. The digits can be overwhelming, and errors can easily occur. Factorials of even moderately large numbers (like 20!) already yield enormous results.
Using Stirling's approximation can convert complex, large-number calculations into manageable multiplications and exponents. The essence of simplifying:

• **Reduce calculation steps**: Break down the complex products
• **Increase computation speed**: Simplify processes
• **Minimize errors**: Fewer steps, fewer mistakes

For instance, calculating \(25!\) directly involves multiplying 25 descending numbers, but Stirling's method turns it into a few multiplications and exponentiations. This is especially useful in fields like physics or computer science, where extremely large numbers are common.

Scientific Calculator

A scientific calculator is a crucial tool for tackling advanced mathematical problems like factorials and approximations. These calculators often have built-in functions for factorials (usually labeled as \(!\)).
Some advanced models can compute Stirling's approximation directly. Key features of a scientific calculator:

• **Factorial Function**: Facilitates quick calculations of \(n!\)
• **Exponential and Logarithmic Functions**: Helpful for parts of Stirling's formula
• **Memory Storage**: Save intermediate values for complex problems

Despite their computing power, understanding the underlying mathematics remains important. Stirling's approximation, for instance, can help verify the results provided by calculators or estimate values in scenarios where access to such tools is limited.